Reading Baez and Stay's "Rosetta Stone," and trying to understand the definition of monoidal category on page 12, I read that a monoidal category requires a natural isomorphism called the associator, assigning to each triple of objects $X, Y, Z \in C$ an isomorphism

and then trying to understand this associator as a natural isomorphism, which is a special case of a natural transformation.

A natural transformation relates a pair of functors $F$ and $F'$ that both map from a category $C$ to a category $D$. To narrow in on the definition of $a$, the associator, I must find two functors and a source category $C$ given an original functor, which, in the case of the associator, is the tensor product $\otimes$, which maps from the cartesian product category $E\times E$ to an original category $E$.

I am imagining that to define $a$ as a natural transformation, I must think of its source category $C$ as $E\times E\times E,$ which should be ok since the cartesian product of categories is evidently associative, and its destination category $D$ as just $E$, getting there via the two different parenthesizations of $X\otimes Y\otimes Z$.

Then the first functor of the natural transformation, $F$ would be something that computes $((X\in E)\otimes(Y\in E))\otimes (Z\in E)$ straight out of the category $E\times E\times E$ into $E$, and the second functor of the natural transformation $F'$ would be something that computes $(X\in E)\otimes((Y\in E)\otimes (Z\in E))$ to go from $E\times E\times E$ to $E$.

This leaves me feeling vaguely queasy, leaning on the associativity of the cartesian product and not really precisely defining the functors $F$ and $F'$, and I'd like to know if I've missed the mark.

EDIT: thinking more about why I feel queasy, it's because I've assumed some kind of projection operator that can "pick" elements from the cartesian product of categories $E\times E\times E$, but such an operator is not immediately obvious from the given definition of the cartesian product of categories on page 11, namely that $C\times C'$ is a category wherein objects are pairs $(X\in C, X'\in C')$, morphisms are applied and composed componentwise, and identity morphisms are defined componentwise. This doesn't, prima facie, give me a way to pick $X$ or $X'$ from $(X, X')$.

2 Answers
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The first functor here is actually a composition $$(\mathcal C\times\mathcal C)\times\mathcal C\xrightarrow{(-\otimes-)\times\text{Id}}\mathcal C\times\mathcal C\xrightarrow{\otimes}\mathcal C\\ ((a,b),c)\mapsto (a\otimes b,\ c)\mapsto(a\otimes b)\otimes c$$ and the second functor is the composition $$\mathcal C\times(\mathcal C\times\mathcal C)\xrightarrow{\text{Id}\times(-\otimes-)}\mathcal C\times\mathcal C\xrightarrow{\otimes}\mathcal C\\ (a,(b,c))\mapsto (a,\ b\otimes c)\mapsto a\otimes( b\otimes c)$$ However, if we want $\alpha$ to be a natural isomorphism between them, we have to consider both parenthesizations of $\mathcal C\times\mathcal C\times\mathcal C$ as equal. But this would be wrong since, formally, they are different. We can resolve this issue by extending the first functor above with the map $\mathcal C\times\mathcal C\times\mathcal C\to(\mathcal C\times\mathcal C)\times\mathcal C$ sending the triple $(a,b,c)$ to the pair $((a,b),c)$ (and similarly for the second functor). It is straightforward to check that this really is a functor. Similarly $(-\otimes-)\times\text{Id}$ is a functor. (Both cases can be seen as applications of the general fact that given functors $F_i$ from a category $\mathcal C$ to the categories $\mathcal D_i$, indexed by some set $I$, there is a unique functor $F:\mathcal C\to\prod_I\mathcal D_i$ such that $p_i\circ F=F_i$ where $p_i:\prod_I\mathcal D_i\to\mathcal D_i$ is the projection.) Finally, $\otimes$ is a functor by hypothesis. So it makes sense to talk about a natural transformation between $(-\otimes-)\otimes-$ and $-\otimes(-\otimes-)$.

The objects of $E\times E\times E$ are the triples $\langle X,Y,Z\rangle$ of objects of $E$. One functor assigns $\langle X,Y,Z\rangle \mapsto (X\otimes Y)\otimes Z$ and the other assigns $\langle X,Y,Z\rangle \mapsto X\otimes (Y\otimes Z)$, and so it is required that these are naturally isomorphic (and that, altogether satisfy the pentagon diagram for quadruples of objects and the unit diagrams).

By the way, this all is much related to the thing about which you wrote

"which should be ok since the cartesian product of categories is evidently associative"

Cartesian product is not associative in the strict sense! Neither that of sets. Because $(A\times B)\times C$ contains pairs of different sets than $A\times (B\times C)$ (and $A\times B\times C$ is a third set). But they are naturally isomorphic. That's a main example.